Deriving the Area of a Sphere

Date: 10/21/2003 at 23:59:24
From: sana
Subject: Deriving the area of sphere 

I know the area of a sphere is 4phi(r^2), but I'm wondering how to 
derive that formula.  I know it should be done in cylindrical 
coordinates, and I'm thinking that the arc of a circle is defined as 
rd(theta) and it's multiplied with rd(phi) to get (r^2)d(theta)d(phi).  
Could you please help explain this?

Date: 10/22/2003 at 02:19:09
From: Doctor Jeremiah
Subject: Re: Deriving the area of sphere 

Hi Sana,

I think the following answer from the Dr. Math archive will tell 
you what you want to know: 

  Surface Area and Volume of a Sphere
    http://mathforum.org/library/drmath/view/55189.html 

Here is another answer not done in spherical coordinates--it's done 
by summing up thin circular sections:

  Deriving the Integral for the Surface Area of a Sphere
    http://mathforum.org/library/drmath/view/52133.html 

And here is one with spherical coordinates:

  Surface Area of a Sphere
    http://mathforum.org/library/drmath/view/51754.html 

And here is one that does it completely differently:

  Surface Area of a Sphere
    http://mathforum.org/library/drmath/view/51761.html 

I found these by searching our archives on these keywords:

   derive area sphere

You can find other similar answers by doing the same search at:

    http://mathforum.org/library/drmath/mathgrepform.html 

I hope this helps.  Let me know if you'd like to talk about this some
more, or if you have any other questions.  Be sure to search our
archives when you have a question--you will often find your answer 
without having to write in and wait for a response.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 10/22/2003 at 03:25:42
From: Doctor Luis
Subject: Re: Deriving the area of sphere 

Hi Sana,

Please reference the diagram at the following url

  Spherical Coordinates
    http://mathworld.wolfram.com/SphericalCoordinates.html 

where theta is the azimuth angle (i.e. the angle measured from the
x-axis if you project the point to the xy-plane), and phi is the polar
angle (i.e. angle measured from the z-axis).  The radius r is a
constant for our sphere.  The angle theta is also known as the
longitude, and the angle phi is also known as the latitude.

Imagine a little patch on the surface of the sphere, created by the
differential angles d(theta) and d(phi).  What is the area of that
"square" patch?  Now, be careful here.

The side from the phi angle is easy, it's just r*d(phi), since it's
just a small differential arc on a great circle on the sphere (if you
look at the diagram, phi is measured on a great circle that has the
z-axis as its diameter).

The side from the theta angle is trickier.  It's still a differential
arc on a circle, so its length is still radius*d(theta), but the 
circle is smaller.  You can see that on the diagram.  The circle at
the equator is largest (with radius r), but circles at higher
latitudes have a smaller radius.  You can see that these circles get
smaller and smaller as you approach the poles.  Do you see that in the
diagram? 

Here is another diagram:

   

As you can see, the radius of that circle is r*sin(phi).  Therefore,
the side of the patch from the d(theta) increment is
(r*sin(phi))*d(theta).

The area of the patch is then

   dA = [r d(phi)] * [r sin(phi) d(theta)]
      = r^2 sin(phi) d(phi) d(theta)

To obtain the whole surface of the sphere, we integrate over all
angles.  The latitude phi ranges from 0 to pi, and the longitude
theta ranges from 0 to 2pi.  Therefore, our area is

       / /            /2pi       /pi
       | |            |          |
   A = | | |dA| = r^2 | d(theta) | sin(phi) d(phi)
       | |            |          |
      / /            / 0        / 0


     = r^2 * (2pi - 0) * (-cos(pi) + cos(0))

     = r^2 * (2pi) * (2)

     = 4pi*r^2

and there you have it.

Let us know if you have any more questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/